Resonant frequency and circumference of spheres

1. Jul 5, 2008

nuby

Does the electromagnetic or acoustic resonant frequency of a hallow spherical object ever equal the circumference of that object? If not what are the formulas to calculate this? Thanks in advance.

2. Jul 5, 2008

Xezlec

A circumference is a distance. A resonant frequency is a frequency. Talking about a distance being equal to a frequency doesn't make sense; it's like asking if your age equals your name.

3. Jul 5, 2008

nuby

True... I think I asked the wrong question..
Anyway, do you know the equation to calculate the resonant frequency of a hollow sphere, and string?

Thanks.

Last edited: Jul 5, 2008
4. Jul 5, 2008

nuby

I meant to ask if wavelength or 1/wavelength is ever equal to the circumference of an spherical object.

5. Jul 5, 2008

f95toli

In which context?
I suggest you google "whispering gallery resonator",although they are no spherical but cylindrical.

6. Jul 5, 2008

nuby

Last edited by a moderator: May 3, 2017
7. Jul 6, 2008

nuby

Does this look right for sphere resonance?

f = v / (2*pi*r)

8. Jul 6, 2008

9. Jul 7, 2008

marcusl

The previous post described EM resonances internal to a conducting sphere. The complementary exterior problem results in scattering of EM waves off of a conducting sphere. This is known as Mie scattering and the solutions and equations are not simple (the scattered waves are combinations of reflections from the front with resonant "creeping waves" that travel over the spherical surface). However resonances begin when $$ka = 1$$, where $$ka$$ is the circumference of the sphere measured in wavelengths.

You can find the mathematical details in Jackson, Classical Electrodynamics, 2nd ed., section 9.13. The plot of how the scattering amplitude changes with circumference is found in every discussion of radar scattering cross section. See, e.g., Fig. 7 in
http://www.tscm.com/rcs.pdf" [Broken]

Last edited by a moderator: May 3, 2017
10. Jul 8, 2008

nuby

A little off topic, but ...
Does the Schumann resonance have more to do with the ionosphere or the molten iron within the earth? According to Wiki, it has to do with the ionosphere and Earth's surface. But why is this:

Schumann resonance equation: c/(2*pi*earth_radius) * sqrt(n*(n+1))

Which comes out to be 7.49 hz ..

But the real Schumann frequency is 7.8 hz ...

Since the upper mantel depth is right around 250,000 meters

Wouldn't c/(2*pi*(earth_radius - upper_mantel_depth)) make sense?
So you get the real Schumann frequency of 7.8hz?

11. Jul 8, 2008

marcusl

The earth's surface looks like a good conductor when viewed on a large scale. Ground is used as the return conductor for AC power transmission, for instance. The wavelength at 7 Hz = 37000 km, so underground water tables contribute to the apparent conductivity, and any local surface features are irrelevant. The earth is largely covered with conductive oceans, furthermore. I doubt that you need to consider the molten core.
EDIT: I just noticed you are talking about a depth of just 250 km. I suppose there might be some penetration that deep, although I still think the crust and water tables are quite conductive.

Last edited: Jul 8, 2008
12. Jul 9, 2008

marcusl

Had a chance to look into Schumann resonances a little. Jackson points out that, of the two boundaries, it is the ionosphere that departs most from a perfect conductor and as a result the resonances vary with daily changes in the ionosphere.

Given these factors, your agreement (7.5 vs. 7.8 Hz, or within 4%) is excellent.